\newproblem{lay:6_2_9}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.2.9}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\mathbf{u}_1=\begin{pmatrix}1\\0\\1\end{pmatrix}$, $\mathbf{u}_2=\begin{pmatrix}-1\\4\\1\end{pmatrix}$, and $\mathbf{u}_3=\begin{pmatrix}2\\1\\-2\end{pmatrix}$.
	Show that the set $\mathcal{B}=\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}$ is an orthogonal basis of $\mathbb{R}^3$. Let $\mathbf{x}=\begin{pmatrix}8\\-4\\-3\end{pmatrix}$.
	Express $\mathbf{x}$ as a linear combination of the $\mathbf{u}_i$'s.
}{
   % Solution
	To check whether $\mathcal{B}$ is orthogonal, we calculate all possible inner products to check if they are 0 or not
	\begin{center}
	  \begin{tabular}{l}
			$\mathbf{u}_1\cdot\mathbf{u}_2=0$\\
			$\mathbf{u}_1\cdot\mathbf{u}_3=0$\\
			$\mathbf{u}_2\cdot\mathbf{u}_3=0$\\
		\end{tabular}
	\end{center}
	So, the set $\mathcal{B}$ is orthogonal. To express $\mathbf{x}$ as a linear combination of the $\mathbf{u}_i$'s we construct the matrix
	$A=\begin{pmatrix}\mathbf{u}_1 &\mathbf{u}_2 &\mathbf{u}_3\end{pmatrix}$ and solve the equation system $A[\mathbf{x}]_{\mathcal{B}}=\mathbf{x}$
	\begin{center}
		$\begin{pmatrix}1 & -1 & 2\\ 0 & 4 & 1 \\ 1 & 1 & -2\end{pmatrix}[\mathbf{x}]_{\mathcal{B}}=\begin{pmatrix}8\\-4\\-3\end{pmatrix} \Rightarrow
		[\mathbf{x}]_{\mathcal{B}}=\begin{pmatrix}1 & -1 & 2\\ 0 & 4 & 1 \\ 1 & 1 & -2\end{pmatrix}^{-1}\begin{pmatrix}8\\-4\\-3\end{pmatrix}=
		   \begin{pmatrix}2.5\\-1.5\\2\end{pmatrix}$
	\end{center}
}
\useproblem{lay:6_2_9}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
